$2)A=\dfrac{\sqrt{x}}{\sqrt{x}-3}(x≥0,x\ne9)$
$⇔\dfrac{\sqrt{x}}{\sqrt{x}-3}=\dfrac{3}{2}$
$⇔2\sqrt{x}=3\sqrt{x}-9$
$⇔\sqrt{x}=9$
$⇔x=81$(thỏa)
Vậy `S={81}`
$3)B=\dfrac{2}{x+\sqrt{x}+1}(x>0)$
$⇔\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{7}$
$⇔14=2x+2\sqrt{x}+2$
$⇔2x+2\sqrt{x}-12=0$
$⇔2(x+\sqrt{x}-6)=0$
$⇔x+\sqrt{x}-6=0$
$⇔(\sqrt{x}-2)(\sqrt{x}+3)=0$
$⇔\sqrt{x}-2=0$ hoặc $\sqrt{x}+3=0$
$⇔x=4$(thỏa) hoặc $\sqrt{x}=-3$(vô lý)
Vậy `S={4}`
$1)P=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}(x≥0)$
$⇔\dfrac{\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{1}{\sqrt{x}}$
$⇔x+\sqrt{x}=\sqrt{x}+2$
$⇔x-2=0$
$⇔x=2$(thỏa)
Vậy `S={2}`
